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个人观点，仅供参考。。。
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谢谢关心，没在巡警了。
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终于给出代码了
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编码习惯问题,深拷贝与浅拷贝
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 ls的很淡定，学习，哈哈 |
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是A& operator=( A &objA )
我刚打掉了一个
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你的代码要是我传一个NULL，不知道是不是会崩溃。，其实如果发现题目有错，完全可以帮他更正的，这样又多拿一分
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哈哈，果然死循环了！！！ 但是没有溢出喃
 [root@localhost root]# vi pediy.cpp #include <iostream> class A { protected: int m_n; public: A():m_n(1) { std::cout << "default ctor" << std::endl; } A(const A& a) { std::cout << "copy ctor" << std::endl; *this = a; } A operator=(const A& a) { std::cout << "operator =" << std::endl; if (this == *a) return *this; m_n = a.m_n; return *this; } }; int main(void) { A a; A b(a); } [root@localhost root]# g++ pediy.cpp [root@localhost root]# ./a.out default ctor copy ctor operator = copy ctor operator = copy ctor operator = copy ctor operator = copy ctor operator = copy ctor operator = copy ctor operator = copy ctor operator = copy ctor operator = ...... |
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打内心佩服VCVC，我继续努力
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貌似LZ不泄于搞这种分析
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修正Bug,没有测试，自己测试
#include <iostream> class A { protected: int m_n; public: A():m_n(1) { std::cout << "default ctor" << std::endl; } A(const A& a) { std::cout << "copy ctor" << std::endl; *this = a; } A operator=(const A& a) { std::cout << "operator =" << std::endl; if (this == *a) return NULL; m_n = a.m_n; return *this; } }; int main(void) { A a; A b(a); } |
